show \[[(p\rightarrow q)\land (q\rightarrow r)]\rightarrow (p\rightarrow r)\] is a tautology, without using truth tables
Seems like transitivity.
-[(p>q)n(q>r)] v (p>r) -(p>q) v -(q>r) v (p>r) -(-pvq) v -(-qvr) v (-pvr) (pn-q) v (qn-r) v -p v r
it is entirely obvious, but somehow i get messed up at the last step
ok so i am fine at \[(p\land \lnot q)\lor (q\land \lnot r)\lor (\lnot p \land r)\]
(pn-q) v (qn-r) v-p vr -------------- (pv-p) n (-qv-p) v (qvr) n (-rvr) T n (-qv-p) v (qvr) n T (Tn-q) v (Tn-p) v (qnT) v (rnT) -q v-p v q v r (-qvq) v-p v r T v ..... = T
Can't you just write that it represents a syllogism?
@amistre, you lost me on the last line
yeah i am trying to do it amistre way
since we got all ors, and a T; the rest dont matter
T or ? or ? or ? = T
that is fine , but what is this (pn-q) v (qn-r) v-p vr ?
oring -p to the first bit and r to the last bit i stacked them vertical; helps me see whats going on
ok that is the line i need to get starting here \[(p\land \lnot q)\lor (q\land \lnot r)\lor (\lnot p \land r)\]
\[(p\land \lnot q)\lor (q\land \lnot r)\lor (\lnot p \lor r)\]
a -> changes to "v" not "n"
p -> r :: -p v r
ah ok so we have \[(p\land \lnot q)\lor (q\land \lnot r)\lor (\lnot p \lor r)\] and then \[(p\land \lnot q)\lor (q\land \lnot r)\lor \lnot p \lor r\] and then we are going to commute to \[\lnot p \lor (p \land \lnot q) \lor (q \land \lnot r) \lor q\]
yep
not a vq, vr
good, and from there its cake walk; hot coals cake walk, but cake walk nonetheless :)
\[(\lnot p \lor p )\land (\lnot p \lor \lnot q)\lor (q \lor r) \land (\lnot r \lor r)\]
i think that is right now yes?
yep; then we can T off the sides and distribute thru the middles which cancels out the ts
god this makes my head spin. thanks
yeah, i blame guass lol
well i tried to do this on the fly and got hopelessly stuck, then i tried it again and got stuck again not on the fly. thanks for straightening it out!
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